AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

Consider the diagram, The dynamic equation for mass m1 is,
T−m1g=m1a ......(I)
The dynamic equation for mass m2 is,
m2g−T=m2a .....(II)
From equation (I) and (II), we get
T−m1g+m2g−T=m1a+m2a
a=

(m2−m1)g

t(m1+m2)

a=

(0.72−0.36)(10)

(0.72+0.36)

a=

3.6

1.08

m∕s2
Substitute values in equation (I),
T−(0.36)(10)=(0.36)(

3.6

1.08

)
T=4.8N
The distance travelled by mass m1 is,
s=ut+

1

2

at2
s=(0)(1)+

1

2

(

3.6

1.08

)(1)2
s=

1

2

(

3.6

1.08

)(1)2
s=

1.8

1.08

The required work done is,
W=T×s
W=(4.8)×(

1.8

1.08

)
W=8J